##How to evaluate an expression with one variable

Let's say we want to evaluate the expression $a + 4$. Well, first we need to know the value of the variable $a$. For example, to evaluate the expression when $\blueD {a = 1}$, we just replace $\blueD a$ with $\blueD 1$:

$\begin{align}
&\blueD a + 4
\\
=&\blueD1 + 4~~~~~~~~\gray{\text{Replace }\blueD{a} \text{ with } \blueD{1}\text{.}}
\\
=&5
\end{align}$

So, the expression $a + 4$ equals $5$ when $a = 1$.

We can just as easily evaluate $a + 4$ when $\blueD {a = 5}$:

$\begin{align}
&\blueD a + 4
\\
=&\blueD5 + 4~~~~~~~~\gray{\text{Replace }\blueD{a} \text{ with } \blueD{5}\text{.}}
\\
=&9
\end{align}$

So, the expression $a + 4$ equals $9$ when $a = 5$.

##Evaluating an expression with multiplication

You might be asked to "Evaluate $3x$ when $x = 5$."

Notice how the number $3$ is right next to the variable $x$ in the expression $3x$. This means "$3$ times $x$". The reason we do this is because the old way of showing multiplication with the symbol $\times$ looks confusingly similar to the variable $x$.

Okay, so now let's solve the problem:

$\begin{align}
&3\blueD x
\\
=& 3 \cdot \blueD5~~~~~~~~\text{Replace }\blueD{x} \text{ with } \blueD{5}\text{.}
\\
=&15
\end{align}$

So, the expression $3x$ equals $15$ when $x = 5$.

New ways to show multiplication

Hold on a second! Did you notice that we wrote "$3$ times $\blueD 5$" as $3 \cdot \blueD 5$ instead of as $3 \times \blueD 5$? Using a dot instead of the symbol $\times$ is another new way of showing multiplication:

$3 \cdot \blueD 5 = 15$

Parentheses can also be used to show multiplication:

$3(\blueD 5) = 15$

Let's summarize the new ways of showing multiplication that we learned.

| Old way | New way

  • | :-: | :-:
    With a variable | $3 \times x$ | $3x$
    Without variable | $3 \times 5$ | $3\cdot 5$ or $3(5)$

##Evaluating expressions where order of operations matter

For more complex expressions, we'll have to be sure to pay close attention to order of operations. Let's take a look at an example:

Evaluate $5 + 3e$ when $\blueD{e=4}$.

$\begin{align}
&5+3\blueD e
\\
=&5 + 3 \cdot \blueD 4~~~~~~~~\gray{\text{Replace }\blueD{e} \text{ with } \blueD{4}\text{.}}
\\
=&5 + 12 ~~~~~~~~\text{\gray{Multiply first (order of operations)}}
\\
=&17
\end{align}$

So, the expression $5 + 3e$ equals $17$ when $e = 4$.

Notice how we had to be careful to think about order of operations when evaluating. A common wrong answer is $\redD{32}$, which comes from first adding $5$ and $3$ to get $8$ then multiplying $8$ by $4$ to get $\redD{32}$.

##Let's practice!

[[โ˜ƒ graded-group-set 1]]

##Challenge problems

[[โ˜ƒ graded-group-set 1]]